Question

(7 + 3sqrt(5))/(3 + sqrt(5)) - (7 - 3sqrt(5))/(3 - sqrt(5)) = a + b * sqrt(5)​

Answer 1

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Answer 2

Concept:

Irrational numbers are created when a radical contains an expression that is not a perfect root. So, in order to rationalize the denominator, all radicals in the denominator must be removed.

Given:

The given equation is $\frac{7+3\sqrt{5} }{3+\sqrt{5} } -\frac{7-3\sqrt{5} }{3-\sqrt{5} }=a+b\sqrt{5}.$

Find:

Find the values of a and b.

Solution:

$\frac{7+3\sqrt{5} }{3+\sqrt{5} } -\frac{7-3\sqrt{5} }{3-\sqrt{5} }=a+b\sqrt{5}$

Rationalize the denominator

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}*\frac{3-\sqrt{5}}{3-\sqrt{5}}-[\frac{7-3\sqrt{5}}{3-\sqrt{5}} *\frac{3+\sqrt{5}}{3+\sqrt{5}}]=a+b\sqrt{5}$

Simplify further

$\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-[\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5}]=a+b\sqrt{5}$

( ∵ $(a+b)(a-b)=a^{2} -b^{2}$ )

$\frac{6+2\sqrt{5}}{4} -[\frac{6-2\sqrt{5}}{4}]=a+b\sqrt{5}$

$\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}=a+b\sqrt{5}$

$\frac{4\sqrt{5}}{4} =a+b\sqrt{5}$

$\sqrt{5}=a+b\sqrt{5}$

Or

$0+1\sqrt{5}=a+b\sqrt{5}$

Equating both sides of the equation

$a=0\\b=1$

Hence the values of a and b are 0 and 1.